The difference in the electronegativity of two atoms,when the percentage ionic character is 19.5%, is________.

Answer

To find the difference in electronegativity of two atoms when the percentage ionic character is 19.5%, we can use the formula that relates ionic character to electronegativity:

### Step-by-Step Solution:

1. **Understand the Formula**: The formula for percentage ionic character (P) is given by:
\[ P = 16 \cdot (x_A - x_B) + 3.5 \cdot (x_A - x_B)^2 \]
where \(x_A\) and \(x_B\) are the electronegativities of atoms A and B, respectively.

2. **Set Up the Equation**: We know that the percentage ionic character is 19.5%. Therefore, we can set up the equation:
\[ 19.5 = 16 \cdot (x_A - x_B) + 3.5 \cdot (x_A - x_B)^2 \]

3. **Let \(d = x_A - x_B\)**: To simplify the equation, let \(d\) represent the difference in electronegativity:
\[ 19.5 = 16d + 3.5d^2 \]

4. **Rearrange the Equation**: Rearranging gives us a standard quadratic equation:
\[ 3.5d^2 + 16d - 19.5 = 0 \]

5. **Solve the Quadratic Equation**: We can use the quadratic formula to solve for \(d\):
\[ d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 3.5\), \(b = 16\), and \(c = -19.5\).

6. **Calculate the Discriminant**:
\[ b^2 - 4ac = 16^2 - 4 \cdot 3.5 \cdot (-19.5) = 256 + 273 = 529 \]

7. **Find the Roots**:
\[ d = \frac{-16 \pm \sqrt{529}}{2 \cdot 3.5} = \frac{-16 \pm 23}{7} \]
This gives us two possible solutions:
\[ d_1 = \frac{7}{7} = 1 \quad \text{and} \quad d_2 = \frac{-39}{7} \quad (\text{not valid since electronegativity difference cannot be negative}) \]

8. **Conclusion**: The difference in electronegativity of the two atoms when the percentage ionic character is 19.5% is:
\[ \boxed{1} \]